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GENERAL  ARTICLE

Whoever said Nothing is Impossible? Three Problems from Antiquity

1. Prologue Shashidhar Jagadeeshan In the history of there have been periodical dis- coveries which have changed the very way we think about mathematics. For example, the discovery of non-Euclidean in the early nineteenth century made us question the link between mathematics and reality. G¨odel’swork, in this century, not only introduced the notion of undecid- ability, but also raised very serious questions about the ax- iomatic foundations of mathematics. In a similar vein the Shashidhar Jagadeeshan discovery in the nineteenth century that certain problems in received his PhD in mathematics are impossible to solve, revealed a totally new mathematics from aspect to mathematics. Yes! that’s right, there are prob- Syracuse University, NY, in 1994. His research lems in mathematics that are impossible to solve. It’s not a interests are in the area matter of being clever (you might be a Gauss or a Ramanu- of ‘representation theory jan) or how hard you work, but given certain constraints of finite dimensional these problems are impossible to solve. What’s more, we algebras’. Following his can prove that this is so! The present article will focus on interest in the questions proving the following three problems of antiquity: raised by J Krishnamurti on life and education, he joined Centre For 1. The duplication of a , or the problem of construct- Learning, a small school ing a cube having twice the volume of a given cube. in Bangalore, in 1995. He is interested in 2. The trisection of an , or the problem of dividing developing a a given arbitrary angle into three parts. mathematics curriculum to help students 3. Squaring a , or the problem of constructing a appreciate the inherent square whose area is equal to that of a given circle. beauty and rigour in mathematics. are impossible to solve; under the constraint that we use only a straight edge and compass. The epilogue will discuss some other impossibilities in mathematics. Before I define the problems more clearly and give an outline of their proofs, I would like to give a brief historical introduction to them. All three problems discussed here have their origins in Greek around the 5th century BC. These problems have

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Box 1.

The following is the translation (taken from, [5]) of an ancient document supposedly written by to King III about the year 240 B.C. :

To King Ptolemy, Eratosthenes sends greetings. It is said that one of the ancient tragic poets represented Minos as preparing a tomb for Glaucos and as declaring, when he learnt that it was hundred feet each way: “Small indeed is the tomb thou hast chosen for a royal burial. Let it be double [in volume]. And thou shalt not miss that fair form if thou quickly doublest each side of the tomb.” But he was wrong. For when the sides are doubled, the surface [area] becomes four times as great and the volume eight times. It became a subject of enquiry among geometers in what manner one might double the given volume without changing shape. And this problem was called the duplication of the cube, for given a cube they sought to double it ... .

become famous because they are impossible to solve within the constraints imposed. The resolution of these problems had to wait for nearly 2200 years, for the invention of ab- stract algebra. In the meantime many great mathemati- cians, amateurs and mathematical cranks have had a go at them without much success. To this day mathematics departments around the world receive letters from people (unfamiliar with the fact that these problems are impossible to solve), claiming to have solved them! Like many other long-standing unsolved problems, these too have inspired the discovery and creation of a lot of mathematics such as that of the conic sections, many cubic and quartic and later on algebraic numbers and group theory. Problem 1 might well have originated in the words of an obscure ancient Greek poet representing the mythical King Minos as being dissatisfied with the size of the tomb erected to his son Glaucos (see Box 1). Minos had ordered the tomb be doubled in size. The poet proceeds to say that Minos claimed this could be accomplished by doubling the length of each side. Obviously neither Minos nor the poet passed high school algebra! Attempts to correct the mathematics of the poet led geometers to study the problem of doubling the cube. There is another story related to this problem, where it is said, that the Delians were instructed by their oracle to double the size of ’s cubical altar, so as to get rid of a certain pestilence. This is why this problem is also referred to as the Delian Problem.

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The origins of Problem 2 are most likely in the Greeks’ at- tempt to construct regular polygons. In the construction of a regular nine-sided polygon, one needs to trisect an an- gle of sixty degrees. In fact this problem has attracted a large number of so called ‘angle trisectors’, who to this day attempt to trisect an angle. The history of Problem 3 is tied up with that of finding the area of a circle, and information about this is found in the Rhind (or Ahmes) Papyrus (one of the best known mathematical manuscripts, purchased in 1858 in Egypt by A Henry Rhind and later acquired by the British Museum). Here it states that the area of a circle is equal to that of a square whose side is the diameter diminished by one ninth. This is of course based on the crude approximation of π to be 3.1604 . . .. Since these problems fascinated many mathe- maticians and amateurs, the problem solvers themselves ac- quired names! For example, in Greek times ‘circle squarers’ were referred to by the term ‘tetragonidzein’, which means to occupy oneself with the quadrature. As you can see these problems come with a very rich history, and have many fascinating anecdotes connected with them, so interested readers can pursue them in standard books on the history of mathematics [1]. The problems were finally settled in the nineteenth century. P Wantzel showed in 1837 that Problems 1 and 2 are impossible to solve and in 1882, Problem 3 was shown to be impossible to solve by F Linde- mann. 2. Ruler and Compass Constructions Anyone who has done high school geometry is familiar with constructions using only a ruler and compass. You may well remember constructions like bisecting an angle, drawing the perpendicular bisector of a given and so on. Apart from some of these easy and obvious constructions it’s amazing what one can do with these tools. For example, one can construct a square having the same area as that of a lune, i.e., a plane figure bounded by two circular arcs! For an excellent account and proof of this fact, I would urge you to read [2]. It is not clear why the Greeks imposed these

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constraints on themselves. Perhaps it was due to the no- tion of that the only ‘perfect’ figures were the straight line and the circle, or it was an intellectual game, with very precise rules. It must be noted here that the Greeks them- selves did not always restrict themselves to these tools and did not hesitate to use other instruments. We will discuss this briefly in the epilogue. So what then are the precise rules involved in constructions using a ruler and a compass? Remember by ruler we mean a ‘straight edge’; one cannot use it to make measurements or transfer lengths, only draw straight lines. If you notice all constructions using ruler and compass start with a given set S = {q1, q2, . . . , qn} of points and we use the ruler and compass in the following two ways:

Operation 1(ruler) Through any two points in S draw a straight line. Operation 2 (compass) Draw a circle, whose centre is a point of S, and whose radius is equal to the distance between some pair of points in S.

We note here that the Greeks preferred a more restricted compass than the one we allow. Their compass was a ‘col- lapsing’ one which when lifted from the paper collapsed, not allowing one to maintain a fixed radius. Their compass al- lowed one to place the compass point at one point of S and the pencil tip at another point of S and draw a circle or an arc. It can be shown that every construction that can be performed by a noncollapsing compass such as ours can be performed by a collapsing one ( see Exercise 5.11, pp.59–60 of [3]). While doing ruler and compass constructions you might have realized that all constructions involve the of a line with another line, a line with a circle or a circle with another circle. How do we formalize these ideas? We start by making two definitions. Definition 2.1. The points of of any two dis- tinct lines or using operations 1 or 2, are said to be

28 RESONANCE  March 1999 GENERAL  ARTICLE constructible in one step from S. A point p is said to be con- p3 structible from S if there is a finite sequence p1, p2, . . . , pn = p of points such that for each i = 1, . . . , n the point pi is p5 constructible in one step from the set S ∪{p1, p2, . . . , pi−1}. p p1 2 Definition 2.2. Let γ be a with absolute value |γ|. Then γ is said to be constructible if we have p two constructible points pi,pj whose distance apart is |γ| 4 units, the starting set S for the points pi, pj being the set Figure 1. Bisecting a line of points{p0, p1} whose distance apart is 1 unit. segment using ruler and Let us now use a simple construction as an example to il- compass. plustrate1 our formal set-up. Suppose we are given two points p1, p2. Let S = {p1, p2}. (Figure 1)

1. Draw the line p1p2 (operation 1); 2. Draw the circle centre p1 of radius p1p2 (operation 2); 3. Draw the circle centre p2 of radius p1p2 (operation 2); 4. Let p3 and p4 be the points of intersection of these circles; 5. Draw the line p3p4 (operation 1); 6. Let p5 be the intersection of the lines p1p2 and p3p4.

Figure 2. Doubling the cube.

Then the sequence of points p3, p4, p5 defines a construction of the midpoint of the line p1p2. We are now ready to see what all this formal stuff looks like when we consider the three problems of antiquity. Doubling the Cube. We are given a cube and asked to construct a cube with double its volume. Given two points p d p p1 and p2 whose distance apart is equal to one side of the 1 2 original cube, our task is then to construct√ two points pi and 3 pj whose distance apart is exactly 2 times the distance between p1 and p2(Figure 2).

Trisecting an Angle. We are given points p1, p2 and p3 which determine the angle and are asked to construct a point pi using only operations 1 and 2, such that angle p1p2pi is exactly one third that of the angle p1p2p3. (Figure 3)

Squaring the Circle. Given two points p1 and p2, whose distance apart is the radius of the circle, construct two points √ p p pi and pj whose distance apart is π times the distance i d j

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between p1 and p2.(Figure 4) 3. How Does Algebra Come into the Picture? We will now show how we convert these problems into the language of algebra and can also sympathize with why the Greeks had so much trouble with them. As mentioned earlier, not only was the word algebra ‘Greek’ to the Greeks, they also did not realise that these problems were impossible to solve. To convert these problems into algebra, the reader will now have to be patient as we build up the required background. For reasons of brevity we can’t Figure 3. Trisecting an give a comprehensive background and would recommend any angle. standard text in algebra say [4]. Moreover, we will not give complete proofs, but will only give a sketch. Many of the definitions and statements of theorems are taken directly from the books referred below. The algebra that is used is based on the work of many great algebraists, like Abel, Galois, Lagrange and more recently Artin. The material is also the beginnings of the beautiful area of mathematics known as . Our aim is to characterize constructible numbers algebraically. Figure 4. Squaring the In other words convert the geometric idea of a constructible circle. number as defined in Definition 2.2 to that of an algebraic one and understand it in the language of algebra. Actually, looked at correctly algebra comes in immediately! Theorem 3.1. The set of constructible numbers is a subfield of R. Furthermore every is constructible √ and if γ is a positive then so is γ. Proof: The first part follows from the well known fact that if α and β are constructible, then so are α + β, αβ and α/β. If we are given a unit length, then we can construct all the rationals. Once a given length γ > 0 is constructed, then it √ is fairly easy to construct (using ruler and compass) γ. 2 By repeated applications of the above theorem, for example,

30 RESONANCE  March 1999 GENERAL  ARTICLE one can prove that the number r √ 4 √ q √ 4 13 + 6 + 1 + 2 7 3 is constructible. The theorem tells us that the class of constructible numbers is large. In fact our aim is to find the ones that are not constructible.√ More specifically, if we are able to double the cube then 3 2 has to be constructible, and if we can square √ a circle π would be constructible. Before we proceed we need to understand the notion of a number being algebraic over a field. The set of constructible numbers is a subset of the set of algebraic numbers, so information about the latter will be useful to understand the former.

Definition 3.1. A number α ∈ C is said to be algebraic over a field F contained in C if there exists a non-zero poly- nomial f(x) ∈ F [X] such that α is a zero of f(x). If no such exists then α is said to be transcendental over F . In other words, if α is algebraic over F there exists a poly- n nomial f(X) = a0 + a1X + . . . + anX whose coefficients a0, a1, . . . an are all in F with at least one of these√ coefficients being non-zero and f(α) = 0. For example 2 is algebraic over Q, since it satisfies the polynomial X2 − 2 = 0. Recall that a polynomial is monic if its leading coefficient is 1.

Definition 3.2. Let α be algebraic over F ⊆ C. Then there exists a unique monic polynomial f(X) over F of least degree such that f(α) = 0 and it is called the of α over F , and is denoted irr (α, F ). √ For example, X2 −2 is the irreducible polynomial of 2 over Q. We denote the degree of irr(α, F ) by deg(α, F ). Definition 3.3. Let F be a field. A non constant polyno- mial f(X) ∈ F [X] is said to be irreducible over F if it has no factors in F [X]. The following theorem connects the notion of the irreducible

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polynomial of a number and that of an irreducible polyno- mial over a field. Moreover given α ∈ C algebraic over F , it gives us a way of computing irr(α, F ).

Theorem 3.2. Let F be a subfield of C and let α ∈ C be algebraic over F . The following conditions on a polynomial f(X) ∈ F [X] are equivalent: (i) f(X) = irr(α, F ). (ii) f(α) = 0 and f(X) is monic irreducible over F . In general it is not easy to find out if a given polynomial is irreducible or not. One powerful criterion is Eisenstein’s criterion (see Box 2 for details). Every time we have a field F embedded in another field K, i.e.,F ⊆ K, we obtain a vector space K over F . The dimension of K over F is denoted by [K : F ]. The field K is called the extension field of F . The next theorem tells us when a given α ∈ K is algebraic over F .

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Theorem 3.3. Let F be a subfield of a field K with doubt [K : F ] = n where n is a positive finite . Then every number α ∈ K is algebraic over F and deg(α, F ) ≤ n. One can extend a given field F by ‘attaching’ a single ele- ment α (not in F , but in K) to F .

Definition 3.4. Let F be a subfield of C and let α be alge- braic over F with deg(α, F ) = n. The extension of F by α is n−1 the set F (α) ⊆ C where F (α) = {b0 +b1α+. . .+bn−1α : b0, b1, . . . , bn−1 ∈ F }. Theorem 3.4. Let F be a subfield of C and let α ∈ C be algebraic over F with deg(α, F ) = n. Then (i) F (α) is a finite dimensional vector space over F with basis {1, α, α2, . . . , αn−1}. The dimension n of F (α) over F as a vector space is denoted by [F (α) : F ]. (ii) F (α) is a field. √ √ For example the set Q ( 2) = {a + b 2 : a, b ∈Q} is a field extension of Q of degree 2. Actually for our purpose we will be mainly interested in extensions of fields of degree 2, the so called quadratic extensions. Now whenever we perform a construction using ruler and compass it is not enough to have an extension of the rationals by a single element; in fact we have to iterate the process to obtain a sequence of extensions. For example to find a suitable extension of Q where r √ 4 √ q √ 4 13 + 6 + 1 + 2 7 3 lives, we need the following sequence of field extensions : √ q √ √ F0 = Q, F1 = F0( 7), F2 = F1( 1 + 2 7), F3 = F2( 6), r q √ √ q√ F4 = F3( 6 + 1 + 2 7), F5 = F4( 13), F6 = F5( 13).

We now discuss briefly what happens if we have a tower of distinct subfields of C, E ⊆ F ⊆ K. We notice that there are three vector spaces that are involved here. F over E, K over F and K over E. It is natural to ask if, say, the dimensions of the vector

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spaces F over E and K over F are finite, what is the di- mension of K over E ? The following theorem gives the answer.

Theorem 3.5. Consider a tower of subfields of C, E ⊆ F ⊆ K. If the vector spaces F over E and K over F have finite dimensions, then so does K over E and [K : E] = [K : F ][F : E].(Figure 5) 4. The Proofs of the Impossibilities at Last ! Figure 5. Tower of subfields of C. We are now almost ready to prove the impossibilities of the three constructions – we just need the following theorem that does the trick! The background material from the previous section was necessary and hopefully sufficient to understand this theorem. We have noticed from the example above that having a sequence of quadratic extensions produces a con- structible number. The question then is : ‘is there any other way to produce constructible numbers ?’. The answer is ‘no’! Theorem 4.1.The following are equivalent :

(i) The number γ is constructible.

(ii) There exists a sequence of positive real numbers γ1, γ2, . . . , γn such that γ ∈ F where F = , 1 1 1 Q √ γ2 ∈ F2 where F2 = F1( γ1), . . . √ γn ∈ Fn where Fn = Fn−1( γn−1), and, finally, √ γ ∈ Fn+1 where Fn+1 = Fn( γn).

Proof: (ii) implies (i) follows from Theorem 3.1 and induc- tion on n. The proof that (i) implies (ii) rests on the fact that to produce a new constructible point one needs the in- tersection of either a line with a line, or a line with a circle or that of a circle with a circle. In each of the above cases at worst one may need extra square roots to describe the new point constructed, in other words a quadratic extension. We illustrate with an example below. 2

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Example 4.2. Start with two points P0 = (0, 0) and P1 = (1, 0) at a distance 1 unit apart. Draw circles with centres at P0 and P1 with radius P0P1 then the√ points of intersection√ P2 and P3 can be computed to be (1/2, 3/2)√and (1/2, − 3/2) respectively involving nothing more than 3. (Figure 6). Corollary 4.3. If γ is constructible, then γ is algebraic and deg(γ, Q)= 2s for some integer s ≥ 0. The proof follows from the above Theorems 4.1 and 3.5. Corollary 4.4. If a real number γ satisfies an irreducible polynomial over Q of degree n, and if n is not a power of 2, then γ is not constructible. This is an immediate consequence of the above corollary. We are now really ready to prove that the three great prob- lems of antiquity are impossible to solve! Theorem 4.5. The cube cannot be duplicated using ruler- and-compass constructions.

Proof:√As discussed earlier, if we could duplicate√ the cube then 3 2 would be constructible. However 3 2 satisfies the

Figure 6.

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monic polynomial X3 − 2 = 0 over Q, which by Eisenstein’s 3 criterion is irreducible over√Q. But the degree of X − 2 = 0 is 3 and by Corollary 4.4, 3 2 is not constructible. Hence it is impossible to duplicate the cube using ruler-and-compass constructions. We now show that the angle π/3 cannot be trisected by a ruler-and- compass construction. Thus there is no ruler- and-compass construction which will provide a trisection of every angle. This leaves the possibility of being able to trisect ‘most’ , with the method failing only for a finite number of angles. However it turns out that once we have established the fact that π/3 is not constructible, it is quite easy to deduce that there are infinitely many angles that cannot be trisected ( see [5], p.106). Theorem 4.6. The angle π/3 cannot be trisected using ruler-and-compass constructions. Proof: If the angle π/3 could be trisected, then the angle π/9 can be constructed. Using a right angled triangle it is easy to see that the length cos(π/9) and hence β = 2 cos(π/9) can be constructed. Recalling the fact that cos(π/3) = 1/2 and the trignometric identity cos(3θ) = 4 cos3(θ) − 3 cos(θ), we see that β satisfies the monic polynomial β3 − 3β − 1 = 0. Now the polynomial f(X) = X3 − 3X − 1 is irreducible over Q, since the polynomial f(X + 1) = X3 + 3X2 − 3 is irreducible over Q, by Eisenstein’s criterion. But once again the degree of f(X) = X3 − 3X − 1 is 3 and by Corollary 4.4, β is not constructible. Hence π/3 cannot be trisected. 2 Theorem 4.7. The circle cannot be squared using ruler- and-compass constructions. Proof: As discussed earlier if we could square the circle, √ then π would be constructible and hence by Corollary 4.3 algebraic. This implies that π is algebraic over Q, but there is a famous Theorem of Lindemann (see [3], Chapter 6, for a proof) which asserts that π is not algebraic over Q. Hence it is impossible to square a circle. 2 Neat, isn’t it?

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5.Epilogue Suggested Reading

In the process of writing this article, I have discovered that [1][1][1] Howard Eves, An Intro-Intro-An I have only touched the tip of the iceberg, both in terms of duction to the history of geometric constructions and in terms of mathematical im- Mathematics, Holt,Holt,, possibilities. Rinehart and Winston, New York, 1976. As far as geometric construction goes, it has been known [2][2][2] William Dunham, since the time of the Greeks that if one relaxed the con- Journey Through Genius,,, dition on the use of ruler and compass alone, then these John Wiley and Sons, three problems could be solved. For example, Problem 1 1990.1990.1990. was solved by the Greeks using a ruler in the form of a right [3][3][3] Ian Stewart, GaloisGaloisGalois TheoryTheoryTheory, Chapman and angle (see p.146 of [6], for details) and using the spiral of Hall, London, 1991. one can square a circle and trisect an angle (see [4][4][4] I N Herstein, Topics in pp. 95–96, of [1] for details). There are even variations with AlgebraAlgebraAlgebra, Blaisdell, New regard to the traditional ruler and compass as tools. If we York,1964.York,1964.York,1964. agree that a line can be drawn if we are given two points on [5][5][5] Arthur Jones, A Sidney it, then it can be shown that all the ruler and compass con- Morris and Kennneth R. structions can be constructed using a compass alone (first Pearson, Abstract Alge- discovered by Georg Mohr in 1672, and later discovered by bra and Famous Impossi- bilitiesbilitiesbilities, Springer-Verlag, and attributed to Lorenzo Mascheroni). It is also known New York, 1991. that all ruler and compass constructions can be performed [6][6][6] Richard Courant and by using a two-edged ruler, whose edges are either Herbert Robbins, What or meet at a point. is Mathematics?, OxfordOxfordOxford University Press, New One of the reasons Gauss is revered so much is that at the age York, 1941.1941.York, of 19, he demonstrated how to construct a [7][7][7] M Rosenlicht, Integra- of 17 sides, using straight edge and compass! (See box item tion in Finite Terms, by Renuka Ravindran in Resonance, Vol.2, No.6, p.62) This American Mathematical after nearly 2000 years of stagnation as far as construction MonthlyMonthlyMonthly,79,963–972, of regular polygons was concerned. Till then the Greeks 1972.1972.1972. [8][8][8] R Hadlock Charles, had constructed regular polygons with 3 sides, 5 sides and in Theory and its m general regular polygons with 2 sides, for m ≥ 2. They also Classical Problems, CarusCarus, knew that if they could construct polygons with r sides and Mathematical Mono- s sides, then they could construct one with side rs. Gauss graphs, No. 19, Mathe- showed further, that a regular polygon of n sides can be matical Association of constructed whenever n is prime and is of the form n = 2k+1 America, 1978. for some integer k ≥ 0. These primes are actually called Fermat primes. You can check that 17 is one such. So from our discussion we see that we can construct a regular polygon of n sides provided it is of the form n = 2i+2 or i n = 2 p1p2 . . . pj, where i ≥ 0 and pi, 1 ≤ i ≤ j, are Fermat

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primes. The question that naturally comes up is, what if n is not of this form. Well! Pierre Wintzel in 1837, proved that if n is not of this form, then it is impossible to construct a regular polygon with n sides. The next famous impossibility comes from the area of solving

polynomial equations with real√ coefficients. We all know −b± b2−4ac the quadratic formula x = 2a , where a 6= 0, is the solution to the equation ax2 + bx + c = 0. If you notice the solution involves only the coefficients of the equation and square roots. Such a solution is referred to by the term solutions by radicals. That is, the solutions are obtained using field operations and extraction of roots alone. Once again it is natural to ask: is there a solution by radicals for a polynomial of arbitrary degree? The answer is yes if the degree is 3 or 4. But for degree 5, Niels Abel showed in 1824 that it is impossible to solve an arbitrary fifth degree equation by radicals alone. Just as in the case of the trisection of an angle, one should note that there may be some fifth degree equation, which does have a solution by radicals, but there is no general formula which takes care of all fifth degree equations. Actually the punchline to this story was given by Ev´ ariste Galois, who showed that in fact for an integer n ≥ 5 it is impossible to solve an arbitrary equation of degree n by radicals. The tools used by Abel and Galois, form what is today known as Galois theory. The last example we choose is from the area of calculus. We all know that no matter how hard we try we are unable to find R ex2 dx in ‘closed’ form, i.e., in terms of functions usually defined in introductory calculus courses. Joseph Li- ouville proved that it is impossible to find R ex2 dx in closed form. In [7] a more modern proof, based on the ideas of Ga- lois theory has been provided. Address for Correspondence So next time you come across the phrase ‘nothing is impos- Shashidhar Jagadeeshan sible’ don’t believe it! Centre for Learning 462, 9th Cross, Jayanagar 1st Block Bangalore 560 011, India Email: [email protected]

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